Question Video: Finding the Integration of a Function Using the Power Rule for Integration with Roots and Negative Exponents | Nagwa Question Video: Finding the Integration of a Function Using the Power Rule for Integration with Roots and Negative Exponents | Nagwa

# Question Video: Finding the Integration of a Function Using the Power Rule for Integration with Roots and Negative Exponents Mathematics • Second Year of Secondary School

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Determine ∫ (−4 √(𝑥) − 5 + (7/𝑥²)) d𝑥.

04:05

### Video Transcript

Determine the integral of negative four root 𝑥 minus five plus seven over 𝑥 squared d𝑥.

So, the first thing we’re gonna want to do here is rewrite our expression if we want to integrate it. I’m gonna rewrite it using some exponent rules. And the first one of these is that if we’ve got the square root of 𝑥, then this is equal to 𝑥 raised to the power of a half. And we’re gonna be able to use this on our first term. And also, we’ve got another exponent rule that we’re gonna use. And that rule is if we have one over 𝑥 to the power of 𝑎, this is equal to 𝑥 to the power of negative 𝑎. And we’re gonna use that on the third term in our expression.

Okay, so now, let’s use these and rewrite our expression. So, when we do, we can see that the expression we’re now trying to integrate is negative four 𝑥 to the power of a half minus five plus seven 𝑥 to the power of negative two. And in order to integrate the expression that we’ve got, we just need to remind ourselves how we integrate individual terms. Well, if we want to integrate 𝑥 to the power of 𝑛 d𝑥, then this is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝑐. So, basically, what we do is we add one to the exponent and divide by the new exponent. And then, we add 𝑐, which is our constant of integration.

So, this means that our first term is gonna be negative four 𝑥 to the power of three over two. And that’s because if you have a half and you add a one, you get one and a half or three over two. And then, this is divided by the new exponent, so divided by three over two. Then, we get minus five 𝑥. And we get that because if we think of five being five 𝑥 to the power of zero, if we raise the exponent by one, we get five 𝑥. And if we divide by the new exponent, we divide by one; it doesn’t change it. And then, we get plus seven 𝑥 to the power of negative one over negative one. And then, finally, add 𝑐 because we can’t forget our constant of integration.

Okay, so now, let’s tidy this up. Well, our first term is gonna be negative eight over three 𝑥 to the power of three over two. And if we look at how we got that, we had negative four divided by three over two. Well, if we divide by a fraction, it’s the same as multiplying by the reciprocal of that fraction. So, that’s negative four multiplied by two over three. Well, negative four multiplied by two is negative eight. So, we get negative eight over three. And then, our second term remains unchanged. So, we have negative five 𝑥. And then, finally, we have minus seven 𝑥 to the power of negative one. And we got that because it just changed the sign of the final term because we’re dividing by negative one. And then, we have our plus 𝑐 on the end.

We could leave the answer like this. This will be totally acceptable. But we’re gonna rewrite it to be in the same format that we had the original expression in. Well, for the first term, what we’re gonna do is we’re going to use our exponent rule we looked at first. And what we’re gonna get is negative eight root 𝑥 cubed over three. So, we can use the first exponent rule to give us root 𝑥. But because we had 𝑥 to the power of three over two, then we use the numerator of that fraction as the exponent of the 𝑥 term within our root. And then, our second term remains unchanged. So, we’ve got minus five 𝑥. And then, finally, for the last term, we use the second exponent rule that we looked at. So, we’re gonna get minus seven over 𝑥. And then, we have our plus 𝑐 on the end.

So, therefore, we can say that the integral of negative four root 𝑥 minus five plus seven over 𝑥 squared d𝑥 is negative eight root 𝑥 cubed over three minus five 𝑥 minus seven over 𝑥 plus 𝑐.

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